Question

The value of $\sum _{n=0}^{100}{i}^{n!}$ equals ( where $i=\sqrt{-1}$ ):

• A. $-1$
• B. $i$
• C. $2i+95$
• D. $96+i$

Answer 1

The correct option is D $96+i$

${\sum }_0^n{i}^{n!}=i^0+i^1+i^2+i^{1\times 2\times 3}+.....+i^{1\times \mathrm{2.....}\times 100}$

We know that , $i^{4n}=1$ , $i^{4n+1}=i$ , $i^{4n+2}=-1$ , $i^{4n+3}=-i$

So starting from n=4 every number will be 1

So our sum shortens to $1+i+(-1)+(-1)+97=96+i$